What is the minimum value of intlimits_0^x {t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $f(x) = \int\limits_0^x {t{e^{ - {t^2}}}} dt$. By the Fundamental Theorem of Calculus,the derivative is $f'(x) = x{e^{ - {x^2}}}$. Setting $f'

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Let $f(x) = \int\limits_0^x {t{e^{ - {t^2}}}} dt$. By the Fundamental Theorem of Calculus,the derivative is $f'(x) = x{e^{ - {x^2}}}$. Setting $f'(x) = 0$,we get $x{e^{ - {x^2}}} = 0$. Since ${e^{ - {x^2}}} 
eq 0$ for all real $x$,we have $x = 0$. Now,find the second derivative: $f''(x) = {e^{ - {x^2}}} - 2{x^2}{e^{ - {x^2}}} = {e^{ - {x^2}}}(1 - 2{x^2})$. Evaluating at $x = 0$,$f''(0) = {e^0}(1 - 0) = 1 > 0$. Since the second derivative is positive at $x = 0$,the function has a local minimum at $x = 0$. The minimum value is $f(0) = \int\limits_0^0 {t{e^{ - {t^2}}}} dt = 0$.

What is the minimum value of $\int\limits_0^x {t{e^{ - {t^2}}}} dt$?

Similar Questions

If the real part of the complex number $(1-\cos \theta+2 i \sin \theta)^{-1}$ is $\frac{1}{5}$ for $\theta \in(0, \pi)$,then the value of the integral $\int_{0}^{\theta} \sin x \,dx$ is equal to:

The value of the integral $\int_{\pi / 6}^{\pi / 2} \left( \frac{1+\sin 2x+\cos 2x}{\sin x+\cos x} \right) dx$ is equal to

$\int_5^{10} \frac{d x}{(x-1)(x-2)} = $

$\int_0^1 |5x - 3| dx = $

$\int_1^{\sqrt{3}} \frac{1}{1 + x^2} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module